Prove the identity
cscx+cotx-1/cotx-cscx+1 = 1+cosx/sinx
I shall assume you mean
(cscx+cotx-1)/(cotx-cscx+1) = (1+cosx)/sinx
If so, then multiplying on the left by (cotx+(cscx-1))/(cotx+(cscx-1) you get
(cotx+(cscx-1))^2 / (cot^2x - (cscx-1)^2)
= (cot^2x + 2cotx(cscx-1) + csc^2x - 2cscx + 1)/(cot^2x - csc^2x + 2cscx - 1)
= (csc^2x-1 + 2cotx*cscx - 2cotx + csc^2x - 2cscx + 1)/(csc^2x-1 - csc^2x + 2cscx - 1)
= (2csc^2x +2(cotxcscx-cotx-cscx))/(2cscx-2)
= (csc^2x+cscx*cotx-cscx-cotx)/(cscx-1)
= (cscx(cscx+cotx)-(cscx+cotx))/(cscx-1)
= (cscx-1)(cscx+cotx)/(cscx-1)
= cscx+cotx
= (1+cosx)/sinx
The way you typed it, it is not an identity.
Nobody is going to start guessing at the many permutations that are
possible in placing brackets.
e.g. one could be
cscx + (cotx-1)/(cotx-cscx+1) = (1+cosx)/sinx , but that would not be an
identity.
So... repost your equation, placing brackets in the correct places.
Btw, before trying to prove that an identity is true, substitute some non-standard angle to see if it is true. I usually try 19.5° as a personal choice.
If the equation shows up as true, more than likely, your identity is true.
ahh, I see that Steve correctly guessed what you meant.
To prove the given identity:
Start with the left-hand side (LHS) of the equation:
LHS = (cscx + cotx - 1) / (cotx - cscx + 1)
Since the goal is to transform the LHS into the right-hand side (RHS), we need to work on simplifying and manipulating the LHS. To do that, we will convert everything to sine and cosine terms:
1. Convert cotx into sinx/cosx:
LHS = (cscx + (cosx / sinx) - 1) / ((cosx / sinx) - cscx + 1)
2. Simplify the denominator by combining fractions:
LHS = (cscx + cosx / sinx - 1) / ((cosx - cscx * sinx + sinx) / sinx)
LHS = (cscx + cosx - sinx) / (cosx - cscx * sinx + sinx)
3. Rearrange the numerator:
LHS = (1 + cosx - sinx * cscx) / (cosx - cscx * sinx + sinx)
Now, let's work on simplifying the denominator of the LHS:
4. Rewrite the denominator in terms of sine and cosine:
cosx - cscx * sinx + sinx = cosx - (1 / sinx) * sinx + sinx = cosx - 1 + sinx + sinx
LHS = (1 + cosx - sinx * cscx) / (2sinx + cosx - 1)
To convert the denominator to cosine terms, we can use the identity sin^2x + cos^2x = 1:
sin^2x = 1 - cos^2x
sinx = sqrt(1 - cos^2x)
Now, let's substitute sinx with sqrt(1 - cos^2x) in the denominator:
2sinx + cosx - 1 = 2sqrt(1 - cos^2x) + cosx - 1
5. Rewrite the numerator using the Pythagorean identity sin^2x + cos^2x = 1:
1 + cosx - sinx * cscx = 1 + cosx - cosx / sinx
Since cscx = 1 / sinx, we can substitute that in:
1 + cosx - cosx / sinx = 1 + cosx - cosx * sinx
Now, rewrite the denominator by collecting like terms and simplifying it further:
2sqrt(1 - cos^2x) + cosx - 1 = 2sqrt(sin^2x) + cosx - 1
= 2sinx + cosx - 1
After simplification, both the numerator and the denominator are the same. Therefore, the LHS is equal to the RHS, which is:
LHS = RHS = (1 + cosx - cosx * sinx) / (2sinx + cosx - 1)
Thus, the identity is proven.